that is same as to right in the expression below:
[tex]\gamma[/tex] = ks/ kt
which [tex]\gamma[/tex] is the ratio of CP and CV heat capacities and ks and kt are the isentropic and isothermal compressibility, respectively.
k can be expressed as below:
kt= -(1/V)(DV/DP)t
and,
ks= -(1/V)(DV/DP)s

once we know these, all you should worry about is to have the ratio of heat capacities ( CP and CV ) out of Tds Equations, and that is by using the first and second ones and the third one is of no use here, there we have:

Tds= CVdT + T(DP/DT)VdV (the first Tds equation)
and,

Tds= CPdT - T(DV/DT)PdP (the second Tds equation)
by putting dS=0 in both equation we wil have,

CV = - [T(DP/DT)VdV]/dT
and,

CP = [T(DV/DT)PdP]/dT
by dividing the two new equations we will have,

CP/CV = - [(DV/DT)PdP]/[(DP/DT)VdV]
more explicitly,

CP/CV = - [(DV/DT)P/(DP/DT)V](DP/DV)?
we should put "S" instead of "?" for started with the assumption ds=0 and that is how got here. That is,

CP/CV = - [(DV/DT)P/(DP/DT)V](DP/DV)S
next time we put dT=0 in the Tds equations (the first and second), then we should have,

ds= (DP/DT)VdV
and,

ds= - (DV/DT)PdP
left sides are equal, so we try putting right sides equal,

(DP/DT)VdV = - (DV/DT)PdP
more explicitly,

(DP/DT)V/(DV/DT)P= - (DP/DV)?
this time we should put "T" instead of "?" for we started with the assumption dT=0. That is,

[(DP/DT)V/(DV/DT)P]= - (DP/DV)T
with a small change we will have,

[(DV/DT)P / (DP/DT)V] = - [ 1/ (DV/DP)T]
from there we will have:

CP/CV = - [- 1 / (DV/DP)T](DP/DV)S
no different from,

CP/CV = (DP/DV)S / (DV/DP)T
================================
one question students may get up to, after working out this problem, is:
Why using TdS equations to solve the problem?
and the answer is not that simple!
you may or maynot find it out later.

One may say it is of the two equal interpretations on thermodynamics, which say:
1- thermodynamics is the science dealing with work, heat and internal energy
2- thermodynamics is the science dealing with entropy and energy.